4-x-x-260-find-the-possible-values-of-x-

Question Number 205238 by necx122 last updated on 13/Mar/24

$$\mathrm{4}^{{x}} \:+\:{x}\:=\:\mathrm{260} \\ $$$${find}\:{the}\:{possible}\:{values}\:{of}\:{x} \\ $$$$ \\ $$

Commented by Ghisom last updated on 13/Mar/24

generally b^x =ax+c let x=−(t/(ln b))−(c/a) e^t t=−((ln b)/(b^(c/a) a)) t=W (−((ln b)/(b^(c/a) a))) x=−(c/a)−(1/(ln b))W (−((ln b)/(b^(c/a) a)))

$$\mathrm{generally} \\ $$$${b}^{{x}} ={ax}+{c} \\ $$$$\mathrm{let}\:{x}=−\frac{{t}}{\mathrm{ln}\:{b}}−\frac{{c}}{{a}} \\ $$$$\mathrm{e}^{{t}} {t}=−\frac{\mathrm{ln}\:{b}}{{b}^{{c}/{a}} {a}} \\ $$$${t}={W}\:\left(−\frac{\mathrm{ln}\:{b}}{{b}^{{c}/{a}} {a}}\right) \\ $$$${x}=−\frac{{c}}{{a}}−\frac{\mathrm{1}}{\mathrm{ln}\:{b}}{W}\:\left(−\frac{\mathrm{ln}\:{b}}{{b}^{{c}/{a}} {a}}\right) \\ $$

Answered by A5T last updated on 13/Mar/24

4^4 =2^8 =256=260−4⇒x=4 when x>4, 4^x +x>256+4=260 when x<4, 4^x +x<260⇒x=4

$$\mathrm{4}^{\mathrm{4}} =\mathrm{2}^{\mathrm{8}} =\mathrm{256}=\mathrm{260}−\mathrm{4}\Rightarrow{x}=\mathrm{4}\: \\ $$$${when}\:{x}>\mathrm{4},\:\mathrm{4}^{{x}} +{x}>\mathrm{256}+\mathrm{4}=\mathrm{260} \\ $$$${when}\:{x}<\mathrm{4},\:\mathrm{4}^{{x}} +{x}<\mathrm{260}\Rightarrow{x}=\mathrm{4}\: \\ $$

Commented by necx122 last updated on 13/Mar/24

can it be solved analytically? I know of Lambert W function but I don't know how to apply it here.

Answered by mr W last updated on 13/Mar/24

4^x =260−x 4^(x−260) ×4^(260) =260−x (260−x)4^(260−x) =4^(260) (260−x)e^((260−x)ln 4) =4^(260) (260−x)ln 4e^((260−x)ln 4) =4^(260) ×ln 4 ⇒(260−x)ln 4=W(4^(260) ×ln 4) ⇒x=260−((W(4^(260) ×ln 4))/(ln 4))=4

$$\mathrm{4}^{{x}} =\mathrm{260}−{x} \\ $$$$\mathrm{4}^{{x}−\mathrm{260}} ×\mathrm{4}^{\mathrm{260}} =\mathrm{260}−{x} \\ $$$$\left(\mathrm{260}−{x}\right)\mathrm{4}^{\mathrm{260}−{x}} =\mathrm{4}^{\mathrm{260}} \\ $$$$\left(\mathrm{260}−{x}\right){e}^{\left(\mathrm{260}−{x}\right)\mathrm{ln}\:\mathrm{4}} =\mathrm{4}^{\mathrm{260}} \\ $$$$\left(\mathrm{260}−{x}\right)\mathrm{ln}\:\mathrm{4}{e}^{\left(\mathrm{260}−{x}\right)\mathrm{ln}\:\mathrm{4}} =\mathrm{4}^{\mathrm{260}} ×\mathrm{ln}\:\mathrm{4} \\ $$$$\Rightarrow\left(\mathrm{260}−{x}\right)\mathrm{ln}\:\mathrm{4}={W}\left(\mathrm{4}^{\mathrm{260}} ×\mathrm{ln}\:\mathrm{4}\right) \\ $$$$\Rightarrow{x}=\mathrm{260}−\frac{{W}\left(\mathrm{4}^{\mathrm{260}} ×\mathrm{ln}\:\mathrm{4}\right)}{\mathrm{ln}\:\mathrm{4}}=\mathrm{4} \\ $$

Commented by necx122 last updated on 13/Mar/24

This is so clear and understandable. Thank you sir.

Commented by necx122 last updated on 13/Mar/24

$${Meanwhile},\:{is}\:{the}\:{Lambert}\:{W}\:{function} \\ $$$${supposed}\:{to}\:{give}\:{us}\:{other}\:{values}? \\ $$$${How}\:{do}\:{we}\:{calculate}\:{for}\:{others}? \\ $$$${What}\:{calculators}\:{do}\:{we}\:{use}\:{for} \\ $$$${computing}\:{the}\:{lambart}\:{W}\:{function}. \\ $$

Commented by mr W last updated on 13/Mar/24